Let $y=\dfrac{x}{\sin(x)}$. $\dfrac{dy}{dx}=$
Solution: $\dfrac{x}{\sin(x)}$ is the quotient of two, more basic, expressions: $x$ and $\sin(x)$. Therefore, $\dfrac{dy}{dx}$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\dfrac{x}{\sin(x)}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(x)\sin(x)-x\dfrac{d}{dx}(\sin(x))}{(\sin(x))^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{1\cdot \sin(x)-x\cdot \cos(x)}{\sin^2(x)}&&\gray{\text{Differentiate }x\text{ and }\sin(x)} \\\\ &=\dfrac{\sin(x)-x\cos(x)}{\sin^2(x)}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{\sin(x)-x\cos(x)}{\sin^2(x)}$ or any other equivalent form.